Messages in this thread | | | Date | Wed, 19 May 2010 12:27:43 +0200 | From | Oleg Nesterov <> | Subject | Re: schedule() && prev/current (Was: [PATCH 3/3] Make get_current() __attribute__((const))) |
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On 05/19, Peter Zijlstra wrote: > > On Tue, 2010-05-18 at 23:22 +0200, Oleg Nesterov wrote: > > > And, looking at this patch I think that schedule() can be simplified > > a little bit. > > > > "sched: Reassign prev and switch_count when reacquire_kernel_lock() fail" > > commit 6d558c3ac9b6508d26fd5cadccce51fc9d726b1c says: > > > > Assume A->B schedule is processing, > > ... > > Then on B's context, > > ... > > prev and switch_count are related to A > > > > How so? switch_count - yes, we should change it. But prev must be > > equal to B, and it must be equal to current. When we return from > > switch_to() registers/stack should be restored correctly, so we > > can do > > What if schedule() got called at a different stack depth than we are > now? > > I don't think we can assume anything about the stack context we just > switched to.
Not sure I understand...
OK. Firstly, we shouldn't worry about the freshly forked tasks, they never "return" from switch_to() but call ret_from_fork()->schedule_tail(), right?
Now suppose that A calls schedule() and we switch to B. When switch_to() returns on B's context, this context (register/stack) matches the previous context which was used by B when it in turn called schedule(), correct?
IOW. B calls schedule, prev == B. schedule() picks another task, prev is saved on B's stck after switch_to(). A calls schedule(), prev == A before context_switch(A, B), but after that switch_to() switches to B's stack and prev == B.
No?
I am looking into the git history now... and I guess I understand why reacquire_kernel_lock() uses current. Because schedule() did something like
prev = context_switch(prev, next); // prev == last
finish_task_switch(prev);
reacquire_kernel_lock(current); // prev != current
Oleg.
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